Problem: Multiply the following rational expressions and simplify the result. $\dfrac{15z-18}{36-25z^2} \cdot \dfrac{25z^2+60z+36}{9z^2+36}=$
Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator, $15z-18$, of the first expression can be factored as $3(5z-6)$ by factoring out $3$. The denominator, $36-25z^2$, of the first expression can be factored as $(6-5z)(6+5z)$ using the difference of squares pattern. The numerator, $25z^2+60z+36$, of the second expression can be factored as $(5z+6)(5z+6)$ using the perfect square pattern. The denominator, $9z^2+36$, of the second expression can be factored as $9(z^2+4)$ by factoring out $9$. Now the product looks as follows: $\dfrac{3(5z-6)}{(6-5z)(6+5z)}\cdot\dfrac{(5z+6)(5z+6)}{9(z^2+4)}$ To find the product of two rational expressions, we multiply across, and simplify. [What's that?] $\begin{aligned} &\phantom{=} \dfrac{3(5z-6)}{(6-5z)(6+5z)}\cdot\dfrac{(5z+6)(5z+6)}{9(z^2+4)} = \dfrac{3(5z-6) \cdot (5z+6)(5z+6)}{(6-5z)(6+5z) \cdot 9(z^2+4)} &\text{Multiply across.}\\\\\\&=\dfrac{\cancel{{3}}\cdot -1\cancel{{(6-5z)}} \cancel{{(5z+6)}}(5z+6)}{ \cancel{{3}}{ \cdot 3}\cancel{{(6-5z)}}\cancel{{(6+5z)}} (z^2+4)} &\text{Cancel out common factors.}\\\\\\\\ &=-\dfrac{5z+6}{3(z^2+4)}\end{aligned}$ Therefore, the simplified form of the product is $-\dfrac{5z+6}{3(z^2+4)}$, which is equivalent to $-\dfrac{5z+6}{3z^2+12}$.